题目链接

  题意：有N个点，下标从0开始，问的是有多少个子树满足给出的H+1个条件，H代表深度（从0开始），每个深度要求对应的结点的个数是大于等于ai的，满足其一就要加入贡献。

  思路：直接Dsu，但是由于重儿子上面的贡献会被直接储存，所以求重儿子的时候，我们需要的是直接加上他的答案即可。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 5e5 + 7;
int N, H, head[maxN], cnt, need[maxN];
ll ans = 0;
struct Eddge
{
    int nex, to;
    Eddge(int a=-1, int b=0):nex(a), to(b) {}
}edge[maxN];
inline void addEddge(int u, int v)
{
    edge[cnt] = Eddge(head[u], v);
    head[u] = cnt++;
}
int siz[maxN], dfn[maxN], tot, rid[maxN], Wson[maxN], deep[maxN] = {0};
void pre_dfs(int u)
{
    siz[u] = 1; dfn[u] = ++tot; rid[tot] = u; Wson[u] = -1;
    int maxx = 0;
    for(int i=head[u], v; ~i; i=edge[i].nex)
    {
        v = edge[i].to;
        deep[v] = deep[u] + 1;
        pre_dfs(v);
        siz[u] += siz[v];
        if(maxx < siz[v])
        {
            maxx = siz[v];
            Wson[u] = v;
        }
    }
}
int now_have[maxN] = {0}, _get[maxN] = {0};
void dfs(int u, bool keep)
{
    for(int i=head[u], v; ~i; i=edge[i].nex)
    {
        v = edge[i].to;
        if(v == Wson[u]) continue;
        dfs(v, false);
    }
    if(~Wson[u])
    {
        dfs(Wson[u], true);
        _get[u] += _get[Wson[u]];
        ans += _get[Wson[u]];
    }
    for(int i=head[u], v, id; ~i; i=edge[i].nex)
    {
        v = edge[i].to;
        if(v == Wson[u]) continue;
        for(int j=dfn[v]; j < dfn[v] + siz[v]; j++)
        {
            id = rid[j];
            now_have[deep[id]]++;
            if(now_have[deep[id]] == need[deep[id]])
            {
                ans ++;
                _get[u]++;
            }
        }
    }
    now_have[deep[u]] ++;
    if(now_have[deep[u]] == need[deep[u]]) { ans ++; _get[u]++; }
    if(!keep)
    {
        _get[u] = 0;
        for(int i=dfn[u], v; i<dfn[u] + siz[u]; i++)
        {
            v = rid[i];
            now_have[deep[v]] = 0;
        }
    }
}
inline void init()
{
    cnt = 0;
    for(int i=0; i<N; i++) head[i] = -1;
}
int main()
{
    scanf("%d%d", &N, &H);
    init();
    for(int i=1, ff; i<N; i++)
    {
        scanf("%d", &ff);
        addEddge(ff, i);
    }
    for(int i=0; i<=H; i++)
    {
        scanf("%d", &need[i]);
        if(!need[i]) ans += N;
    }
    pre_dfs(0);
    dfs(0, false);
    printf("%lld\n", ans);
    return 0;
}